calculate as a limit of sum limit 1 to 2 (3x2-1) - Maths - Application of Integrals

Question

calculate as a limit of sum limit 1 to 2 (3x2-1)

Ajanta Trivedi · Accepted Answer

since the integral as a sum of limits can be written as

$\int_{a}^{b} f (x) ⅆ x = \lim_{n \to \infty} h \sum_{r = 0}^{n - 1} f (a + r h) w h e r e h = \frac{b - a}{n}$ ............(1)

here we have to find the $\int_{1}^{2} (3 x^{2} - 1) ⅆ x$

therefore a=1 and b=2, , $h = \frac{2 - 1}{n} \Rightarrow n h = 1$

$\int_{1}^{2} f (x) ⅆ x = \lim_{n \to \infty} h \sum_{r = 0}^{n - 1} f (1 + r h) w h e r e h = \frac{1}{n} \int_{1}^{2} f (x) ⅆ x = \lim_{n \to \infty} \frac{1}{n} \sum_{r = 0}^{n - 1} f (1 + r h) \int_{1}^{2} f (x) ⅆ x = \lim_{n \to \infty} \frac{1}{n} [f (1) + f (1 + h) + f (1 + 2 h) + ........... + f (1 + (n - 1) h)]$ .............(2)

since $f (x) = 3 x^{2} - 1$

$f (1) = 3 \cdot 1^{2} - 1 f (1 + h) = 3 \cdot (1 + h)^{2} - 1 f (1 + 2 h) = 3 \cdot (1 + 2 h)^{2} - 1 a n d s o o n ...... f (1 + (n - 1) h) = 3 \cdot [1 + (n - 1) h]^{2} - 1$

therefore :

$\int_{1}^{2} (3 x^{2} - 1) ⅆ x = \lim_{n \to \infty} \frac{1}{n} [(3 \cdot 1^{2} - 1) + {3 \cdot (1 + h)^{2} - 1} + {3 \cdot (1 + 2 h)^{2} - 1} + ...... + {3 \cdot (1 + (n - 1) h)^{2} - 1}] = \lim_{n \to \infty} \frac{1}{n} [3 n \cdot 1^{2} - 1 \cdot n + 3 h^{2} {1^{2} + 2^{2} + ....... + (n - 1)^{2}} + 3 \cdot 2 h {1 + 2 + ..... + (n - 1)}] = \lim_{n \to \infty} \frac{1}{n} [3 n - n + 3 h^{2} \frac{(n - 1) n (2 (n - 1) + 1)}{6} + 6 h \frac{(n - 1) n}{2}] = \lim_{n \to \infty} \frac{1}{n} [2 n + n h^{2} \frac{(n - 1) (2 n - 1)}{2} + 3 n h (n - 1)] = \lim_{n \to \infty} [2 + h^{2} \frac{(n - 1) (2 n - 1)}{2} + 3 h (n - 1)] = \lim_{n \to \infty} [2 + \frac{(n h - h) (2 n h - h)}{2} + 3 (n h - h)]$

put nh=1, and h=0 therefore the given integration becomes:

$\int_{1}^{2} (3 x^{2} - 1) ⅆ x = 2 + \frac{1 \cdot 2}{2} + 3 (1 - 0) = 2 + 1 + 3 = 6$